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Question

Physics Question on Combination of capacitors

Two capacitors of 3μF3 \, \mu F and 6μF6 \, \mu F are connected in series and a potential difference of 900 V is applied across the combination. They are then disconnected and reconnected in parallel. The potential difference across the combination is

A

Zero

B

100 V

C

200 V

D

400 V

Answer

400 V

Explanation

Solution

Q=CVQ = CV, Here Q is a constant C1V \therefore C \propto \frac{1}{V} C1C2=V2V1\therefore \frac{C_{1}}{C_{2}} = \frac{V_{2}}{V_{1}} 36=V2V1\Rightarrow \frac{3}{6} = \frac{V_{2}}{V_{1}} V1=2V2\Rightarrow V_{1} = 2 V_{2} Also V1+V2=900VV_{1} + V_{2} = 900 \,V 2V2+V2=900V\therefore 2V_{2} + V_{2} = 900\, V V2=300V V_{2} = 300 \,V and V1=600VV_{1} = 600\, V Common potential V=C1V1+C2V2C1+C2V = \frac{C_{1}V_{1} + C_{2}V_{2}}{C_{1} + C_{2}} =3×106×600+6×106×3003×106+6×106= \frac{3\times10^{-6} \times600 + 6 \times10^{-6} \times300}{3 \times10^{-6} + 6 \times10^{-6} } =1800+1800×1069×106=36009= \frac{1800 +1800 \times 10^{-6}}{9\times 10^{-6}} = \frac{3600}{9} =400V= 400 \,V